Show that height of the cylinder of greatest volume which can be inscribed in a right circular cone of height h and semi vertical angle α is one-third that of the cone and the greatest volume of cylinder istan2 α.
The given right circular cone of fixed height (h) and semi-vertical angle (α) can be drawn as:
Here, a cylinder of radius R and height H is inscribed in the cone.
Then, ∠GAO = α, OG = r, OA = h, OE = R, and CE = H.
r = h tan α
Now, since ΔAOG is similar to ΔCEG, we have:
Now, the volume (V) of the cylinder is given by,
And, for, we have:
∴By second derivative test, the volume of the cylinder is the greatest when
Thus, the height of the cylinder is one-third the height of the cone when the volume of the cylinder is the greatest.
Now, the maximum volume of the cylinder can be obtained as:
Hence, the given result is proved.